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3 years ago

What is 9,235,009 rounded to the nearest hundred thousand

Mathematics

2 answers:

Rudiy273 years ago

8 0

Rounded to the nearest hundred thousand would be 9,200,000.

LiRa [457]3 years ago

5 0

9,240,000 is the answer for your question.

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Daria sells televisions. She earns a fixed amount for each television and an additional $15 if the buyer gets an extended warran

Genrish500 [490]

Answer: $70

Step-by-step explanation:

11 x 15 = 165 - this is the amount she gets for all 11 of the extended warranties

935 - 165 = 770 - this is the amount she gets for all 11 of the televisions

770 / 11 = 70 - this is the amount she gets for 1 television

6 0

3 years ago

Find the sum. 2x^2 +26x+ 156=0

Mila [183]

Answer:

-13

Step-by-step explanation:

2x²+26x+156=0

x²+13x+78=0

x²+13x=-78

x²+13x+169/4=-143/4

(x+13/2)²=-143/4

x+13/2=(i√143)/2 or x+13/2=-(i√143)/2

x=(i√143)/2-13/2 or x=-(i√143)/2-13/2

(i√143)/2-13/2+(-(i√143)/2)-13/2=(i√143-13-i√143-13)/2=-26/2=-13

5 0

3 years ago

Read 2 more answers

Which equations are equivalent to y=2/3x-4 when written in slope-intercept form? Check all that apply.

Vlada [557]

Answer:

B and C

Step-by-step explanation:

The reason is on the picture.

4 0

4 years ago

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Due to a packaging error, 4 cans labeled diet soda were accidentally filled with regular soda and placed in a 12 pack carton of

Sophie [7]

Answer:

Probability that both cans were regular soda = $\frac{1}{11}$

Step-by-step explanation:

Probability = $\frac{Desired outcome}{Total possible outcomes}$

We are given 12 total number of cans; 4 cans have been accidentally filled with diet soda.

Probability that first can is a regular soda:

Outcome that first can is a regular soda will give us the number of regular soda available which are 4

Using formula of probability

Total possible outcomes are, n(total) = 12

Desired outcome: 4 (cans of regular soda)

P(1st can) = $\frac{4}{12}$ = $\frac{1}{3}$

Probability that 2nd can is a regular soda:

<em>As we have already taken a can of regular soda from the pack, the total soda in the pack now 11 and the regular soda left are 3.</em>

Total possible outcomes are, n(total) = 11

Desired outcome: 3 (cans of regular soda as one has already been taken)

P(2nd can) = $\frac{3}{11}$

Probability that both cans are regular soda:

P(both) = P(1st can) × P(2nd can)

= $\frac{1}{3} * \frac{3}{11}$

= $\frac{1}{11}$

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4 years ago

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Simplify.

Alenkasestr [34]

Answer is <span>C) 1/4^2
hope it helps</span>

7 0

3 years ago